What is 1/4 ln|x|-1/8 ln|x^2+4|+C ?

The solution to 1/4 ln|x|-1/8 ln|x^2+4|+C is ln((|x|^{1/4}|x^2+4|^{7/8})/(|x^2+4|))+C

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simplify 1/4 ln|x|-1/8 ln|x^2+4|+C

simplify

\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4 + C

ln \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣} + C

Solution steps

\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4 + C

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{4} ln ∣ x ∣ = ln (∣ x ∣^{\frac{1}{4}})

= ln (∣ x ∣^{\frac{1}{4}}) - \frac{1}{8} ln x^{2} + 4 + C

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{8} ln x^{2} + 4 = ln (x^{2} + 4^{\frac{1}{8}})

= ln (∣ x ∣^{\frac{1}{4}}) - ln (x^{2} + 4^{\frac{1}{8}}) + C

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (∣ x ∣^{\frac{1}{4}}) - ln (x^{2} + 4^{\frac{1}{8}}) = ln (\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}})

= ln (\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}}) + C

Simplify

\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}} : \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣}

= ln \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣} + C

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What is 1/4 ln|x|-1/8 ln|x^2+4|+C ?
The solution to 1/4 ln|x|-1/8 ln|x^2+4|+C is ln((|x|^{1/4}|x^2+4|^{7/8})/(|x^2+4|))+C